Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/26123)

The rewrite relation of the following TRS is considered.

0(1(2(1(x1)))) 1(2(1(1(0(1(2(0(1(2(x1)))))))))) (1)
0(1(2(1(x1)))) 1(2(1(1(0(1(2(0(1(2(0(1(2(x1))))))))))))) (2)
0(1(2(1(x1)))) 1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))) (3)
0(1(2(1(x1)))) 1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))))))))) (4)
0(1(2(1(x1)))) 1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))))))))) (5)
0(1(2(1(x1)))) 1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))))))))))))))) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
1(2(1(0(x1)))) 2(1(0(2(1(0(1(1(2(1(x1)))))))))) (7)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))) (8)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))) (9)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))) (10)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))))))) (11)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))))))))) (12)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
1#(2(1(0(x1)))) 1#(x1) (13)
1#(2(1(0(x1)))) 1#(2(1(x1))) (14)
1#(2(1(0(x1)))) 1#(1(2(1(x1)))) (15)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))))))))) (16)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))))) (17)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))) (18)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))) (19)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))) (20)
1#(2(1(0(x1)))) 1#(0(2(1(0(1(1(2(1(x1))))))))) (21)
1#(2(1(0(x1)))) 1#(0(1(1(2(1(x1)))))) (22)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.